Power Series Solutions to the Legendre Equation

Transcription

1 Power Series Solutions to the Legendre Equation Department of Mathematics IIT Guwahati

2 The Legendre equation The equation (1 x 2 )y 2xy + α(α + 1)y = 0, (1) where α is any real constant, is called Legendre s equation. When α Z +, the equation has polynomial solutions called Legendre polynomials. In fact, these are the same polynomial that encountered earlier in connection with the Gram-Schmidt process. The Eqn. (1) can be rewritten as [(x 2 1)y ] = α(α + 1)y, which has the form T (y) = λy, where T (f ) = (pf ), with p(x) = x 2 1 and λ = α(α + 1).

3 Note that the nonzero solutions of (1) are eigenfunctions of T corresponding to the eigenvalue α(α + 1). Since p(1) = p( 1) = 0, T is symmetric with respect to the inner product (f, g) = 1 1 f (x)g(x)dx. Thus, eigenfunctions belonging to distinct eigenvalues are orthogonal.

4 Power series solution for the Legendre equation The Legendre equation can be put in the form where y + p(x)y + q(x)y = 0, p(x) = 2x α(α + 1) and q(x) = 1 x 2 1 x, if x Since 1 (1 x 2 ) = n=0 x 2n for x < 1, both p(x) and q(x) have power series expansions in the open interval ( 1, 1). Thus, seek a power series solution of the form y(x) = a n x n, x ( 1, 1). n=0

5 Differentiating term by term, we obtain y (x) = na n x n 1 and y = n(n 1)a n x n 2. n=1 n=2 Thus, 2xy = 2na n x n = 2na n x n, and (1 x 2 )y = = = n=2 n=1 n=0 n(n 1)a n x n 2 n(n 1)a n x n n=2 (n + 2)(n + 1)a n+2 x n n=0 n(n 1)a n x n n=0 [(n + 2)(n + 1)a n+2 n(n 1)a n ]x n. n=0

6 Substituting in (1), we obtain (n+2)(n+1)a n+2 n(n 1)a n 2na n +α(α+1)a n = 0, n 0, which leads to a recurrence relation Thus, we obtain a n+2 = (α n)(α + n + 1) a n. (n + 1)(n + 2) α(α + 1) a 2 = a 0, 1 2 (α 2)(α + 3) 2 α(α 2)(α + 1)(α + 3) a 4 = a 2 = ( 1) a 0, 3 4 4!. n α(α 2) (α 2n + 2) (α + 1)(α + 3) (α + 2n 1) a 2n = ( 1) a 0. (2n)!

7 Similarly, we can compute a 3, a 5, a 7,..., in terms of a 1 and obtain (α 1)(α + 2) a 3 = a (α 3)(α + 4) 2 (α 1)(α 3)(α + 2)(α + 4) a 5 = a 3 = ( 1) 4 5 5!. n (α 1)(α 3) (α 2n + 1)(α + 2)(α + 4) (α + 2n) a 2n+1 = ( 1) (2n + 1)! Therefore, the series for y(x) can be written as y 1 (x) = 1 + y(x) = a 0 y 1 (x) + a 1 y 2 (x), where n=1 ( 1)n α(α 2) (α 2n+2) (α+1)(α+3) (α+2n 1) (2n)! x 2n, and y 2 (x) = x + n=1 ( 1)n (α 1)(α 3) (α 2n+1) (α+2)(α+4) (α+2n) (2n+1)! x 2n+1. a 1 a 1

8 Note: The ratio test shows that y 1 (x) and y 2 (x) converges for x < 1. These solutions y 1 (x) and y 2 (x) satisfy the initial conditions y 1 (0) = 1, y 1(0) = 0, y 2 (0) = 0, y 2(0) = 1. Since y 1 (x) and y 2 (x) are independent, the general solution of the Legendre equation over ( 1, 1) is y(x) = a 0 y 1 (x) + a 1 y 2 (x) with arbitrary constants a 0 and a 1.

9 Observations Case I. When α = 0 or α = 2m, we note that α(α 2) (α 2n + 2) = 2m(2m 2) (2m 2n + 2) = 2n m! (m n)! and (α + 1)(α + 3) (α + 2n 1) = (2m + 1)(2m + 3) (2m + 2n 1) = (2m + 2n)!m! 2 n (2m)!(m + n)!. Then, in this case, y 1 (x) becomes y 1 (x) = 1 + (m!)2 (2m)! m ( 1) k (2m + 2k)! (m k)!(m + k)!(2k)! x 2k, k=1 which is a polynomial of degree 2m. In particular, for α = 0, 2, 4(m = 0, 1, 2), the corresponding polynomials are y 1 (x) = 1, 1 3x 2, 1 10x x 4.

10 Note that the series y 2 (x) is not a polynomial when α is even because the coefficients of x 2n+1 is never zero. Case II. When α = 2m + 1, y 2 (x) becomes a polynomial and y 1 (x) is not a polynomial. In this case, y 2 (x) = x + (m!)2 (2m + 1)! m ( 1) k (2m + 2k + 1)! (m k)!(m + k)!(2k + 1)! x 2k+1. k=1 For example, when α = 1, 3, 5 (m = 0, 1, 2), the corresponding polynomials are y 2 (x) = x, x 5 3 x 3, x 14 3 x x 5.

11 The Legendre polynomial Let P n (x) = 1 [n/2] ( 1) r (2n 2r)! 2 n r!(n r)!(n 2r)! x n 2r, r=0 where [n/2] denotes the greatest integer n/2. When n is even, it is a constant multiple of the polynomial y 1 (x). When n is odd, it is a constant multiple of the polynomial y 2 (x). The first five Legendre polynomials are P 0 (x) = 1, P 1 (x) = x, P 2 (x) = 1 2 (3x 2 1) P 4 (x) = 1 8 (35x 4 30x 2 + 3), P 5 (x) = 1 8 (63x 5 70x x).

12 Power Series Solutions to the Legendre Equation Figure : Legendre polynomial over the interval [ 1, 1] RA/RKS MA-102 (2016)

13 Rodrigues s formula for the Legendre polynomials Note that (2n 2r)! (n 2r)! x n 2r = d n dx n x 2n 2r and 1 r!(n r)! = 1 n! ( n r ). Thus, P n (x) in (2) can be expressed as P n (x) = 1 d n [n/2] ( n ( 1) r 2 n n! dx n r r=0 ) x 2n 2r. When [n/2] < r n, the term x 2n 2r has degree less than n, so its nth derivative is zero. This gives P n (x) = 1 d n n ( ) n ( 1) r x 2 n n! dx n r 2n 2r = 1 d n 2 n n! dx (x 2 1) n, n r=0 which is known as Rodrigues formula.

14 Properties of the Legendre polynomials P n (x) For each n 0, P n (1) = 1. Moreover, P n (x) is the only polynomial which satisfies the Legendre equation and P n (1) = 1. (1 x 2 )y 2xy + n(n + 1)y = 0 For each n 0, P n ( x) = ( 1) n P n (x). 1 P n (x)p m (x)dx = 1 { 0 if m n, 2 2n+1 if m = n.

15 If f (x) is a polynomial of degree n, we have f (x) = n c k P k (x), where k=0 c k = 2k f (x)p k (x)dx. It follows from the orthogonality relation that 1 1 g(x)p n (x)dx = 0 for every polynomial g(x) with deg(g(x)) < n. *** End ***